Problem: Which integral gives the arc length of the curve $y=\cos(x)$ over the interval $[0, \pi]$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $ \int_{0}^{\pi} \sqrt{1+\sin^2x}\,dx$ (Choice B) B $ \int_{0}^{\pi} \sqrt{1+\cos x}\,dx$ (Choice C) C $ \int_{0}^{\pi} \sqrt{1+\sin x}\,dx$ (Choice D) D $ \int_{0}^{\pi} \sqrt{1+\cos^2x}\,dx$
Solution: The arc length $L$ of the curve $y=f(x)$ over the interval $[a, b]$ is $ L=\int_a^b\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}\, dx$. First, calculate $dy/dx$. $\begin{aligned} y &= \cos x\\ \\ \dfrac{dy}{dx} &= -\sin x \end{aligned}$ Now apply the arc length formula on the interval $[0, \pi]$ and simplify the integral. $\begin{aligned} L &= \int_{0}^{\pi} \sqrt{1+\left(-\sin x\right)^2}\,dx \\\\ &= \int_{0}^{\pi} \sqrt{1+\sin^2x}\,dx \end{aligned}$